Galileo & Einstein - second thoughts

Discussion in 'Physics & Math' started by c'est moi, May 12, 2002.

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  1. Prosoothus

    Prosoothus Registered Senior Member

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    James R,

    You are correct, the laser beams would not curve in an object moving at a constant speed when you compare them to an absolute frame of reference. But when you look at the laser beams from the objects point of view, they will appear to be curving. The "relative" curving of the laser beams can be used to determine the absolute velocity of the object.

    The targets would be fixed to the object at a predetermined length from the lasers. Note: The whole device would be able to be rotated in order to confirm that the lengths to the target are accurate.


    Tom
     
    Prosoothus, May 27, 2002
    #61

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  3. James R

    James R Just this guy, you know? Staff Member

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    Tom,

    <i>...the laser beams would not curve in an object moving at a constant speed when you compare them to an absolute frame of reference. But when you look at the laser beams from the objects point of view, they will appear to be curving.</i>

    No, they wouldn't curve in the object's frame of reference either.

    <i>The targets would be fixed to the object at a predetermined length from the lasers. Note: The whole device would be able to be rotated in order to confirm that the lengths to the target are accurate.</i>

    If the targets and lasers are both fixed to the object, then no changes in arrival times of the lasers will occur, provided the object continues to move at constant speed.

    You could use this device to measure acceleration, but not "absolute" velocity.
     
    James R, May 27, 2002
    #62

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  5. Prosoothus

    Prosoothus Registered Senior Member

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    James R,

    You're contradicting yourself. Think about the example of the observer dropping a penny in the train. To the observer in the train, the penny falls straight down, but to the observer outside, the penny curves. The exact opposite would occur if the observer shot a laser beam at the floor of the train. To the observer in the train, the laser beam would curve, but to the observer outside, the laser beam would go straight.

    Let me explain it in another way: Say you are travelling at 90% of c and you're holding a laser in your hand. To the right of you there is a target about a 100 meters away. The target is fixed with you, and is travelling at the same speed you are.

    If you shot the target with you're laser, you would miss the target completely because by the time the laser beam reaches the target, the target would have moved. Even though the laser beam travelled in a straight line, to you it would appear that the laser beam curved. In other words, in the absolute frame of reference, the laser beam follows a straight path, but in your frame of reference, the laser beam would appear to curve. The amount of "percieved" curvature is directly proportional to the speed you are travelling.

    If I shoot a laser beam at a target that is moving away from the laser beam, it will take longer for the laser beam to hit the target than it would if the target was stationary. If I shoot a laser beam at a target that is moving towards the laser beam, the laser beam will hit the target sooner than if the target was stationary. The above would occur even if the I was travelling at the same speed as the target, because the speed of the laser beam is always constant, regardless of the speed of the person shooting it.

    Tom
     
    Prosoothus, May 27, 2002
    #63

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  7. thed

    thed IT Gopher Registered Senior Member

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    Tom

    If what you say above had any resemblance to reality then particles in accelerators could never interact. Take two electrons travelling parallel to each at 0.9c in an accelerator. They exchange virtual photons (travelling at c) to mediate the electrostatic force. If you where correct the virtual photon could never hit the other electron as the 'target' had moved on and no forces could be communicated.

    You're example is trivially falsified.
     
    thed, May 27, 2002
    #64
  8. Prosoothus

    Prosoothus Registered Senior Member

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    Thed,

    Let me remind you that particles in particle accelerators are very close together. This is why they still react to each other at high speeds. In other words, the "virtual" photon exchange happens faster than the particles can move out of the way.

    Are you suggesting that the virtual photons are traveling faster than light?? Or are you saying that the speed and direction of virtual photons are relative to the frame of reference of the particles that are producing them???


    Tom
     
    Prosoothus, May 27, 2002
    #65
  9. thed

    thed IT Gopher Registered Senior Member

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    tom

    Now I get it, a photon won't hit a perpendicular target because light bends except when they are too close. And you're evidence for this astounding insight is?

    How the photon did you infer that I implied virtual photons travel faster than light or are relative to the FOR producing them. Again, learn what is meant by 'the speed of light is invariant'.
     
    thed, May 27, 2002
    #66
  10. Prosoothus

    Prosoothus Registered Senior Member

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    Thed,

    Thanks for avoiding my question.

    Like usual, I give my idea, you claim that it's wrong without indicating why it's wrong or giving me your explanation.

    How convenient for you.

    Note: Let me remind you, I don't post messages on sciforums to argue with people. I'm after the truth, just like I think you are. When I post an idea and someone believes that it is wrong, I would appreciate if that person explains that:

    a) My facts are wrong
    or
    b) My conclusion, based on my facts are wrong.

    Unfortunately, again, you provided neither.

    Tom
     
    Prosoothus, May 28, 2002
    #67
  11. James R

    James R Just this guy, you know? Staff Member

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    Hi Tom,

    <i>You're contradicting yourself.</i>

    No I'm not. You're just not understanding.

    <i>Think about the example of the observer dropping a penny in the train. To the observer in the train, the penny falls straight down, but to the observer outside, the penny curves. The exact opposite would occur if the observer shot a laser beam at the floor of the train. To the observer in the train, the laser beam would curve, but to the observer outside, the laser beam would go straight.</i>

    No. To both observers, the path of the laser beam would be straight. If the observer in the train shot the laser beam towards the floor, the beam would be a vertical line from his point of view. An outside observer would see a slanted line, but not a curved one.

    <i>Let me explain it in another way: Say you are travelling at 90% of c and you're holding a laser in your hand. To the right of you there is a target about a 100 meters away. The target is fixed with you, and is travelling at the same speed you are.

    If you shot the target with you're laser, you would miss the target completely because by the time the laser beam reaches the target, the target would have moved.</i>

    No. If the target is travelling along with you, if you aim at the target you will hit it, no matter how far away. Sure, the target moves, but the light has horizontal motion as well, which means it keeps pace with the target.

    <i>Even though the laser beam travelled in a straight line, to you it would appear that the laser beam curved.</i>

    No, you'd see it travel in a straight line.

    <i>In other words, in the absolute frame of reference...</i>

    There is no absolute frame of reference.

    <i>If I shoot a laser beam at a target that is moving away from the laser beam, it will take longer for the laser beam to hit the target than it would if the target was stationary.</i>

    Yes, but not if the target is a fixed distance away from the source of the beam, like you said it was before.

    <i>The above would occur even if the I was travelling at the same speed as the target, because the speed of the laser beam is always constant, regardless of the speed of the person shooting it.</i>

    Wrong. If the target is a constant distance away from the source, the travel time for the light will be constant, regardless of whether the source and target are moving or not.
     
    James R, May 28, 2002
    #68
  12. c'est moi

    c'est moi all is energy and entropy Registered Senior Member

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    damned

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    I had answered your post James R, which took me quite some time. When I hit the "reply" button, I was suddenly asked to login even though I was already. Instead of going back and save the content of my post, I logged in, suspecting nothing, and I got taken to a non-specified page. I went back with my browser and my post was gone and wasn't posted either.
    I decided afterwards to type it again, and in a little different way, somehow the same thing happens. Now I have found the problem (it's my PC) but I don't feel like going through it again .... maybe in a few days. This really frustrating.
     
    c'est moi, May 28, 2002
    #69
  13. Prosoothus

    Prosoothus Registered Senior Member

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    James R,

    Your right and wrong at the same time. The light would move horizontal and forward at the same time, but if the target was moving very fast, the light couldn't catch up because the light would have to travel forward and horizontal, while the target would only have to move forward.

    Check out this link I found:

    http://www.physics.wustl.edu/~visser/physics-216/notes-light-clock.html

    It demonstrates a device that would give the absolute speed of an object. If you count the amount of times the laser bounces off the two mirrors in this device, you can get the absolute speed of an object. The faster the object would go, the fewer times the laser beam would bounce of the mirrors per second.(Of course, you would need three of these, one for each dimension)

    I guess this device would not only prove that there is an absolute frame of reference, but it would also measure the absolute speed of any object. So much for relativity!!

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    Tom
     
    Prosoothus, May 28, 2002
    #70
  14. c'est moi

    c'est moi all is energy and entropy Registered Senior Member

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    In fact, it's all very logic. Light's speed is constant so it can be used as an absolute FOR. Imagine that light wouldn't move at all, then anyone would see that it could be used as abs. FOR, but now it is like the fastest moving thing we know of, which changes the situation, but it remains the same: constant speed means FOR which is independant, hence absolute
    James R, you may come up with objections against how to built such a device but I'm sure such a device can be built
    however, as long as no one has built one and tested it, we both can't be sure
     
    c'est moi, May 28, 2002
    #71
  15. Gil_W

    Gil_W HU-Hybrid Registered Senior Member

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    Hmmmmmmmmm

    I recommend you to read the book:
    <B>"The elegant Universe" by Brian Greene.</B>

    Groove on.

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    Gil_W, May 28, 2002
    #72
  16. James R

    James R Just this guy, you know? Staff Member

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    Tom,

    <i>Your right and wrong at the same time. The light would move horizontal and forward at the same time, but if the target was moving very fast, the light couldn't catch up because the light would have to travel forward and horizontal, while the target would only have to move forward.</i>

    The light will always "catch up", since in any frame of reference the target, which has mass, must travel slower than the speed of light.

    <i>Check out this link I found...</i>

    Did you check it out? It supports relativity. In fact, it is used to derive a relativistic formula.

    <i>If you count the amount of times the laser bounces off the two mirrors in this device, you can get the absolute speed of an object. The faster the object would go, the fewer times the laser beam would bounce of the mirrors per second.(Of course, you would need three of these, one for each dimension)</i>

    No, that's wrong. In the frame of reference of the mirrors, there will be no variation in the number of times light bounces backwards and forwards, regardless of speed.

    <i>I guess this device would not only prove that there is an absolute frame of reference, but it would also measure the absolute speed of any object. So much for relativity!!</i>

    So much for relativity if you were right, but fortunately for relativity, you're wrong.


    c'est moi:

    You're still confused about reference frames. A reference frame involves two things: <b>space</b> (or distance) and <b>time</b>. In order to measure a <b>speed</b>, we need to have a pre-defined spacetime reference frame. Given a reference frame, any object then has a particular speed. Change the reference frame and, usually, the speed of an object changes.

    For example, when you're driving in your car, the speed of the car in the reference frame of the road might be 60 km/hr. In your reference frame, sitting in the car, the car's speed is zero.

    The funny thing about light is that its speed is the same in all reference frames. The only way that is possible is if different observers' ideas of space and time are different depending on their frame of reference - an idea which leads us automatically to the theory of relativity.


    Gil_W:

    Thanks. I've read it.
     
    James R, May 30, 2002
    #73
  17. Prosoothus

    Prosoothus Registered Senior Member

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    James R,

    Read the link again.

    If the speed of light is constant, and if you increase the horizontal velocity of the light in the device by moving it, the vertical velocity of the light must slow down in order for the light's speed to remain constant.

    If you increase the horizontal velocity of light without decreasing the vertical velocity, the light would be travelling faster than c.

    This concept doesn't take a lot of thought, it's common sense.

    Tom
     
    Prosoothus, May 30, 2002
    #74
  18. Prosoothus

    Prosoothus Registered Senior Member

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    Just one more thing:

    The relativistic properties of frames of reference are not the result of there being no absolute frame of reference, but the result of the absolute frames of reference cancelling each other out.

    Tom
     
    Prosoothus, May 30, 2002
    #75
  19. James R

    James R Just this guy, you know? Staff Member

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    Tom,

    <i>If the speed of light is constant, and if you increase the horizontal velocity of the light in the device by moving it, the vertical velocity of the light must slow down in order for the light's speed to remain constant. If you increase the horizontal velocity of light without decreasing the vertical velocity, the light would be travelling faster than c.</i>

    Yes. I agree with that.

    <i>The relativistic properties of frames of reference are not the result of there being no absolute frame of reference, but the result of the absolute frames of reference cancelling each other out.</i>

    How? Explain.
     
    James R, May 31, 2002
    #76
  20. Adam

    Adam §Þ@ç€ MØnk€¥ Registered Senior Member

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    Absolute frames of reference cancelling each other out? Multiple and opposing absolute frames of reference?
     
    Adam, May 31, 2002
    #77
  21. Prosoothus

    Prosoothus Registered Senior Member

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    James R, and Adam,

    Let me explain:

    In a relative frame of reference:

    Vr=V1-V2

    where Vr is the relative velocity of object 1 compared to object 2
    V1 is the relative velocity of object 1
    V2 is the relative velocity of object 2

    If the observer is object 1 then the observer can give the value of 0 to V1 in order to determine the relative speed of object 2 compared to himself/herself.

    However if you assume that there is an absolute velocity(a) that the entire frame of reference is traveling at, you get the formula:

    V=(V1+a)-(V2+a)

    As you can see Vr from the first formula is equal to V in the second formula. As you can also see, the absolute velocities of the observer and object cancel each other out in the second formula, regardless of the actual value of a.

    Tom
     
    Prosoothus, May 31, 2002
    #78
  22. James R

    James R Just this guy, you know? Staff Member

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    Yes, that's right, Tom. Therefore, the concept of a is superfluous. There is no absolute reference frame.

    Also, you have to realise that v1 and v2 must be relative to something as well, unless they are meaningless.
     
    James R, May 31, 2002
    #79
  23. Prosoothus

    Prosoothus Registered Senior Member

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    James R,

    You're wrong.

    If you dropped a rock on your foot, you would be able to determine it's acceleration without knowing the mass of the rock(ignoring air resistance, of course).

    In this case the mass of the rock in the two formulas

    F=m1*m2*g/d^2

    A=F/m1

    cancel each other out.

    This DOES NOT mean that the rock doesn't have mass.

    Tom
     
    Prosoothus, May 31, 2002
    #80
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