# light is a measure of mass not velocity

Discussion in 'Physics & Math' started by ada, Feb 27, 2001.

1. ### AlexGLike nailing Jello to a treeValued Senior Member

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Another, more general question about it. If the final derived equation does not generate a divide by zero when v=c, but one of the steps in the derivation does, does that invalidate the final equation when v=c? :scratchin:

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3. ### TachBannedBanned

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Let's look at the derivation:

$x'=\gamma(v)(x+vt)$
$t'=\gamma(v)(t+vx/c^2)$

are both valid only if $v<c$

With this in mind, differentiate the above:

$dx'=\gamma(v)(dx+vdt)$
$dt'=\gamma(v)(dt+vdx/c^2)$

We can divide the first expression by the second one only if $v<c$

$w(u,v)=\frac{dx'}{dt'}=\frac{v+dx/dt}{1+v/c^2 dx/dt}=\frac{u+v}{1+uv/c^2}$

While v<c there is nothing stopping us from making $u->c$ (the limit exists). When $u->c$ you get $w->c$

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5. ### TachBannedBanned

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Yes, see above. More correctly, it is valid for $v<c$ and $u->c$.
As you correctly pointed out in your earlier post, you cannot attach a reference frame to a photon, so you cannot have $v=c$ (but you can have $v->c$ with $v<c$).

Last edited: Sep 14, 2011
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7. ### NietzscheHimselfBannedBanned

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But like Pete said, the equation is from a reference point at rest. Which in reality nothing is at rest, and in reality neither photon is at rest.

8. ### PeteIt's not rocket surgeryRegistered Senior Member

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So... you mean no, right?
You are arguing that it is not valid when both u and v are equal to c.

Last edited: Sep 15, 2011
9. ### PeteIt's not rocket surgeryRegistered Senior Member

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"At rest" is just as real as "here".

Photons are never at rest, but anything else can certainly be considered to be at rest.

10. ### TachBannedBanned

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Correct. In math terms $w(v=c,u=c)$ is ill-defined, courtesy of its derivation.
Nevertheless, the limit $w(v->c,u=c)->c$ is well-defined.

11. ### AlexGLike nailing Jello to a treeValued Senior Member

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Are you saying that v = c is defined at the limit?

Trout, PLEASE, can you just say yes or no, without showing the math. Trust me, I'll believe you.

12. ### TachBannedBanned

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Your sentence does not make much sense, I am saying that $v=c$ is not allowed (there is no frame moving at c). That is, $w(c,c)$ is not defined. On the other hand, $w->c$ when $u->c$ and $v->c$.
In other words, the function$w(u,v)$ has a limit when $u->c$ and $v->c$.

13. ### NietzscheHimselfBannedBanned

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$w(u,v)=\frac{u+v}{1+uv/c^2}$

$w(0,2c)=\frac{0+2c}{1+(2*0)/c^2}$

$w(0,2c)=\frac{2c}{1}$

$w(0,2c)=2c$

....

14. ### TachBannedBanned

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You don't know what you are doing.

15. ### NietzscheHimselfBannedBanned

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Why not? If we assume one photon is at rest wouldn't we assume the other is moving at 2c?

16. ### PeteIt's not rocket surgeryRegistered Senior Member

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No. You calculated that 2c by simply adding the two individual speeds together:
w = u + v
2c = c + c

This is the intuitive galilean way to add parallel velocities.

But in practice, it doesn't work if either of u or v is not tiny compared to c.
The whole point of the given equation (relativistic addition of parallel velocities) is that:
• if you measure two objects moving at speeds u and v in opposite directions,
• then you change your reference frame to consider one of them to be at rest (ie you use clocks and rulers at rest with respect to one object),
• then you will measure the other object to be moving at:
w = (u+v)/(1 + uv/c^2)

So that equation suggests that if you assume one photon to be at rest, then you should assume the other photon to be moving at c.

(There is the technical issue that you can't have clocks and rulers at rest with a photon, but that's a side point).

17. ### AlphaNumericFully ionizedRegistered Senior Member

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You can't define a Lorentz boost which takes you from a frame where a photon is moving to one where the photon is at rest. This is because the Lorentz transformation doesn't alter the 4-vector norm but something stationary has 4-vector norm of -1 but something moving at the speed of light has 4-vector norm 0.

Suffice to say that there exists no frame where an observer with mass sees any massive or massless object moving at a speed higher than c.

18. ### NietzscheHimselfBannedBanned

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So the jury is still out on the maximum speed observed by two massless particles.

19. ### TachBannedBanned

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What gives you this bright idea?

20. ### NietzscheHimselfBannedBanned

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These two explanations suggest something else.

21. ### PeteIt's not rocket surgeryRegistered Senior Member

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The jury says that a massless particle can't measure the speed of anything, because they can't have clocks at rest with them.

But we can easily avoid this side issue, NH... replace your photons with electrons.

Consider two electrons moving apart from each other. You measure one moving to the left at 0.9c, the other moving to the right at 0.9c, so you measure the distance between them to be growing at 1.8c.

But what does one electron measure for the other's speed? Hint - it's not 1.8c.

u=0.9c
v=0.9c
w = ?

22. ### NietzscheHimselfBannedBanned

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W=.9944

W=(.8,1)

1+.8/(1+1*.8/c^2)=1

It is the same difference in speed and the calculated difference is .0066

And w=(.1,.7) as opposed to w=(.4,.4)=.6896 is an even larger difference (.058) which is a fairly significant difference.

It just doesn't logically sit right because we could actually be moving .3 towards an object and read a different speed as a point directly centered between them. Which of course is completely irrelevant to the speed the object itself observes as the speed of the other object (.8)

23. ### TachBannedBanned

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Wrong.

No, you can't even apply a simple formula. Try again, what is $w(0.9,0.9)$?