Relativistic limit?

1. Apr 11, 2007

bernhard.rothenstein

Consider the following transformation equations for the momentum and mass of a tardyon
p=gp'(1+V/u')
m=gm'(1+Vu'/cc)
Question: What happens if we take in both sides the limit for u' going to c.
Has that some physical meaning? Thanks for the answer.
soft words and hard arguments

2. Apr 12, 2007

Meir Achuz

Thank you. That is a good demonstration why using the 4-vector (E,p)
is better than u, which has complicated transformation properties.
You also have shown an example of how using m instead of E/c^2 can lead to confusion.
In terms of E and p, taking u' to c is trivial. (Nothing happens.)

3. Apr 13, 2007

bernhard.rothenstein

Thanks for your participation on my thread. As you can see I have made some minor changes in the presentation of the transformation equations where u and u' are the OX(O'X') components of the speed of the same tardyon.
a. Using (1) and (2) we should not use the transformation of speeds. It was used deriving them.
b. Taking the mentioned limit we obtain from a mathematical point of view
p(c)=gp'(c)(1+V/c) (1)
m(c)gm'(c)(1+V/c) (2)
or
E(c)gE'(c)(1+V/c) (3)
and from a physical point of view we have obtained the transformation equations for the momentum and the energy of the photon.
Has that some physics behind it? Some thing happens!:rofl: soft words and hard arguments

4. Apr 13, 2007

Meir Achuz

"Has that some physics behind it?"
I think it is just elementary algebra.

5. Apr 15, 2007

bernhard.rothenstein

simple mathematics or some physics as well

Thanks. Consider the addition law of parallel velocities
u=[u'+V]/[1+u'V/cc]
and take there the limit for u=u'=c in order to obtain c=c.
Consider the relativistic Doppler shift formula in the acoustic wave and take there the limit when the propagation velocity becomes equal to c in order to obtain the Doppler shift formula in the electromagnetic wave.
Is that simple mathematics or it reflects some physical facts?
Is it correct to postulate:
If you have in your hands formulas which account for the behaviour of a tardyon take there the limit u=u'=c in order to obtain the corresponding equations that account for the behaviour of the photon.
soft words and hard arguments

6. Apr 16, 2007

Mentz114

The physical content of such formulae would imply that photons can be created by accelerating matter. This is not true. Photons are born photons and are not transformed matter.

7. Apr 16, 2007

bernhard.rothenstein

photon tardyon

Thanks. It is a fact that formulas for tardyons become formulas for photons. Is there an explanation for that?
There are mnemonic rules in physics. Could be my statement such a rule?

8. Apr 16, 2007

Mentz114

What equations do is certainly of interest to mathematicians. We have no evidence of matter being accelerated to photons. The idea does not hold water because speed is relative.

Last edited: Apr 17, 2007
9. Apr 16, 2007

MeJennifer

A particle that has mass can never reach the speed of light not even in the limit. In fact there is no limit.

Convert the speed to rapidity and you can clearly see that there is no limit.

10. Apr 17, 2007

Mentz114

Hi MJ,
if a rocket with enough fuel accelerates away from me at 1g, then after a certain time won't it reach c relative to me ? If it shone a laser back at me I would see it shift down the energy spectrum until it blinked out. I suppose for the signal to truely reach zero would take an infinite time on my clock, which bears you out.

11. Apr 17, 2007

Gib Z

Enough Fuel? Remember as we get closer to the speed of light, our mass increases making it harder to accelerate. In this endless process of gaining more mass and accelerating, you will never actually reach c.

12. Apr 17, 2007

Meir Achuz

The answer to this is yes, but that is trivial algebra and not a hard argument.
But, in terms of E and p, it is still simpler.

"Consider the relativistic Doppler shift formula in the acoustic wave and take there the limit when the propagation velocity becomes equal to c in order to obtain the Doppler shift formula in the electromagnetic wave."
What do you mean by "the relativistic Doppler shift formula in the acoustic wave"?

13. Apr 17, 2007

Mentz114

No.

'...as you approach c' is nonsense if YOU DON'T SAY WHAT FRAME IS MEASURING THE VELOCITY.

Do you think there's an absolute reference frame ? If you were on the spaceship would you detect an increase in your own mass ?

14. Apr 17, 2007

bernhard.rothenstein

Thanks for the first "yes" answer from you.
IMHO Doppler shift in acoustic wave is
f=gf'(1+V/u') (1)
whereas in the electromagnetic wave it is
f=gf'*1+V/c) (2)
and again (1) becomes (2) for u=u'=c.
Yes or not?
The problem is not " trivial argument" but "why"?
Regards

15. Apr 17, 2007

Mentz114

Any comparison between the acoustic wave and the EM wave is naive and unfruitful. Sound is a longitudinal wave in a compressible medium and nothing like EM radiation.

16. Apr 17, 2007

bernhard.rothenstein

acoustic and electromagnetic wave

C.Moller The Theory of Relativity, Clarendon Press Oxford 1972 chapter 2.9in order to see how acoustic and electgromagnetic waves could be related and how the comparison between them could lead to de Broglie.
What I mentioned is the coincidence between the equations which describe the Doppler shift in the two waves for u=u'=c.
soft words and hard arguments

17. Apr 17, 2007

Mentz114

Yes, my reply is too radical. They are both waves after all, and have that much in common. I do not have access to the book you mention so no comment on that.

18. Apr 17, 2007

robphy

19. Apr 17, 2007

Mentz114

Moller (1952) is available. I'll have a look at it when I get the time, because I enjoy 'old' physics texts.

Thanks for the reference.

20. Apr 18, 2007