Really? Tell us exactly what that means, Einstein? How do you interpret that article?
Uhuh...
Propellantless propulsion, apparently.
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Really? Tell us exactly what that means, Einstein? How do you interpret that article?
Uhuh...
Try this interpretation: The photon has no mass, and we have proven that this conjecture is true to with 10[sup]-51[/sup] grams (i.e., to within an incredibly small experimental uncertainty).
doesnt mean it doesnt have mass. I mean think of all the neutrino particles passing through Earth...and fool would say they have no mass...yet there are Japanese and American neutrino capturing heavy water facilities which do trap neutrinos and are able to study these particles. When it comes to numbers...these so called "massless" particles...actually have some mass.
Neutrinos are fermions and are not considered massless, unlike photons.
I understand ducks. If it walks like a duck and quacks like a duck... But oh no, it clearly isn't a duck because Q defines ducks as being yellow and made out of rubber.
I understand mass too. And some people who don't, just don't want to. It's not as if they study my explanation and point out where I'm mistaken. They just don't want to know, and they use cheapshot insults to try and cover this up. It vexes me that people who pretend to be rational open-minded scientists can behave this way.
I understand mass too. And some people who don't, just don't want to. It's not as if they study my explanation and point out where I'm mistaken. They just don't want to know, and they use cheapshot insults to try and cover this up. It vexes me that people who pretend to be rational open-minded scientists can behave this way.
You will notice that light from the sun can have a pushing effect. This tells you off the bat that energy can be used as a system of repulsion.
With matter as a condensed form of energy, matter can be expanded to make large amounts of energy. Just how much expansion you get out of it is how good your propulsion is.
At the moment, we have a very low expansion rate. Rocket fuel just will not cut it.
The form of energy that is produced is less important for space travel. So long as you direct the output you will get thrust. The problem of propulsion is not what works as thrust, that is too easy, it is meaningful expansion of a fuel that can get you over long distances and is storable in usable amounts.
With matter as a condensed form of energy, matter can be expanded to make large amounts of energy. Just how much expansion you get out of it is how good your propulsion is.
At the moment, we have a very low expansion rate. Rocket fuel just will not cut it.
The form of energy that is produced is less important for space travel. So long as you direct the output you will get thrust. The problem of propulsion is not what works as thrust, that is too easy, it is meaningful expansion of a fuel that can get you over long distances and is storable in usable amounts.
Farsight:
You understand the equation
$$E^2 = m^2 c^4 + p^2 c^2 = (mc^2)^2 + (pc)^2$$
right?
Now. Apply that to a photon for me.
In particular, tell me what value you use for "m" in that equation, when you're dealing with a photon. And do you always use the same value of "m", or not?
You understand the equation
$$E^2 = m^2 c^4 + p^2 c^2 = (mc^2)^2 + (pc)^2$$
right?
Now. Apply that to a photon for me.
In particular, tell me what value you use for "m" in that equation, when you're dealing with a photon. And do you always use the same value of "m", or not?
No problem James. If we look at the equation what we're seeing is a rest mass component, plus a velocity component:
$$E^2 = (mc^2)^2 + (pc)^2$$
For a photon the rest mass m is taken to be zero, so we drop the first portion:
$$E^2 = (pc)^2$$
We can simplify this to:
$$E = pc$$
Then with the loss of a little momentum we can use our photon in pair production to create an electron and a positron, which means $$pc$$ is now reconfigured to the first term:
$$E = mc^2$$
The rest mass is nowadays considered to be simply mass, by definition, though you will sometimes hear the phrase "invariant mass". It is therefore separated from "relativistic mass". IMHO this tends to act as a barrier to any consideration of the localisation of momentum into inertia.
$$E^2 = (mc^2)^2 + (pc)^2$$
For a photon the rest mass m is taken to be zero, so we drop the first portion:
$$E^2 = (pc)^2$$
We can simplify this to:
$$E = pc$$
Then with the loss of a little momentum we can use our photon in pair production to create an electron and a positron, which means $$pc$$ is now reconfigured to the first term:
$$E = mc^2$$
The rest mass is nowadays considered to be simply mass, by definition, though you will sometimes hear the phrase "invariant mass". It is therefore separated from "relativistic mass". IMHO this tends to act as a barrier to any consideration of the localisation of momentum into inertia.
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...allrighty than. Why would solar sails work than? huh?
Photons are made of energy, which has momentum, which is transferred to the sails, hence accelerates them. No mass required.
That's what we've been trying to tell you all along.
So, your problem is understanding 'relativistic mass?'
Then publish that paper and wait for the Nobel prize, because nobody, and I really mean nobody, understands mass.
God.
LOL. I do. And you don't want to.
Did you miss what I said there Q? Here's a reminder:
I understand mass. And some people who don't, just don't want to. It's not as if they study my explanation and point out where I'm mistaken. They just don't want to know, and they use cheapshot insults to try and cover this up. It vexes me that people who pretend to be rational open-minded scientists can behave this way.
All: I'm on holiday for a week, and might be out of contact for the duration.
Not much of an answer to James' question {what is photon's mass you assume?} as the photon can create many different pairs. Easier would be a muon pair, harder would be a proton/ antiproton pair. If it has lots of enegy, the photon may continue to "live" and create other pairs by lossing just the required ENERGY, NOT MASS, of the pair when doing so. Please try again.
It was a plenty fine answer, and you're just nitpicking. An electron has energy of 511keV as does a positron, so you need to start with a gamma photon of 1022keV. Just a tad more in fact.
If you want to understand mass, read MASS EXPLAINED:
http://www.sciforums.com/showthread.php?t=61557.
If you don't, don't. No problem.
If you do, thank you. If you find my explanation wanting and can point out the flaw in my explanation, please accept my grovelling humble apologies in advance. But if you can't or won't, don't try to make up for it by telling me I don't understand mass. That just isn't scientific.
If you want to understand mass, read MASS EXPLAINED:
http://www.sciforums.com/showthread.php?t=61557.
If you don't, don't. No problem.
If you do, thank you. If you find my explanation wanting and can point out the flaw in my explanation, please accept my grovelling humble apologies in advance. But if you can't or won't, don't try to make up for it by telling me I don't understand mass. That just isn't scientific.
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I did not tell you that (although it is obviously true). What I said was you were not answering James's question as there are many differ masses that the pair produced could have produced. Way do you select the electron positron pair mass to assign to the photon (instead of the easier muon pair or harder proton pair)? I.e. Why not any of the three different muons and it anti-muon for a lower mass? Why not the proton and the anti-proton masses?
YOU HAVE NO THEORY (Theories require backing by experiments and mathematical predictions.)
YOU HAVE NOT answered either James or my simple question.
Why should we read long crackpot garbage when you can not even answer simple questions?
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