# A limit problem in SR

1. Nov 14, 2007

### bernhard.rothenstein

Consider the limit of the function

mcc/sqrt(1-uu/cc)
for u=c. (m rest mass, u speed of the particle with rest mass m relative to a given inertial reference frame.

Please tell me what is its physical meaning. Do you see there a connection with the energy of a photon?

2. Nov 15, 2007

### CompuChip

Here are some hints:
$$\frac{m c^2}{\sqrt{1 - \frac{u^2}{c^2}}} = \gamma m c^2 = E_0$$
does that equation look familiar? What happens to $\gamma \text{ as } u \to c$?
What is the energy of a photon?

Now please show us some attempt (please try thinking for yourself before asking us to do it for you).

Last edited: Nov 15, 2007
3. Nov 15, 2007

### bernhard.rothenstein

limit

4. Nov 15, 2007

### robphy

When taking limits, there are usually constraints under which the limit is taken... often they are implicit and obvious. For this question, can you make those constraints explicit?

For instance, What is m? (i now see it above) and (more importantly) is it held fixed while taking this limit?

Last edited: Nov 15, 2007
5. Nov 15, 2007

### CompuChip

Isn't it?
If I made a mistake, please tell me -- it happens sometimes (actually, it happens all the time).

6. Nov 15, 2007

### bernhard.rothenstein

relativistic limit

Cut the index zero at the last E.

7. Nov 15, 2007

### bernhard.rothenstein

relativistic limit

Thanks.
In short my problem is:
W=mcc/sqrt(1-uu/cc)-mcc
with m for the rest mass.
Apply it in the case of a photon (u=c, m=0).
W(photon)=0/0.
Is it correct to extend that result to
W(photon)=hniu
taking into account that the energy of a photon is kinetic in its essence?
Please consider that all I say are questions and not statements.

I

8. Nov 15, 2007

### pmb_phy

Consider how one usually derives that expression, i.e. from the Work-Energy Theorem. For those who are rusty on this I worked it out for relativity on one of my web pages. I'm sure the math could have been streamlined better and will do so when I find a shorter/cleaner derivation. And advice on how to do that is welcome!

The derivation starts with a particle at rest, such as a charged particle which is initially at rest in the inertial frame S. A force is then applied to the particle, e.g. an electric field is established, and the particle begins to accerate, the final velocity then being v (or u as you refer to it as). This derivation assumes that the particle can be at rest and that it can accelerate. Neither of which apply to a photon. I recall a limit process in a book by Thorne et al. See - http://www.pma.caltech.edu/Courses/ph136/yr2002/chap01/0201.2.pdf

See page 16 right after Eq. (1.31)

The question asked is not a homework problem and as such we don't need to see any attempt by Bernhard to respond to this question. I believe Bernhard is seeking the impressions others have on this concept. Bernhard is a very sharp man. In fact he was a physicist professor before he retired.

Pete

9. Nov 16, 2007

### bernhard.rothenstein

limit in special relativity

Thanks Pete
My problem has a more general character.
If I start with the formula which accounts for the longitudinal Doppler effect involving a machine gun which emits bullets at a constant period the bullets hitting a moving target at a period T' then
T'=(1+V/u)/sqrt(1-VV/cc) (1)
where u stands for the speed of the bullet and V for the speed of the target.
If I replace the bullets with light signals (u=c) (1) becomes
T'=(1+V/c)/sqrt(1-VV/cc)
which accounts for the Doppler effect in the optical domain. Do you think that is by chance or there is some physics behind.
In general how does special relativity sccount for the transition from the bullet to the photon?
I learned long time ago that "Natura non facit saltus" ensuring smooth transitions.
Regards

10. Nov 16, 2007

### CompuChip

OK, it could have been a homework question, and in general I like to see what the OP already did/tried. For one, it gives me an impression of the level of the poster, so that I don't go around explaining too basic things to someone who long knows them, or start throwing stuff at him/her which is far above the level required. So basically, precisely to avoid this situation

... in which case my response was probably not very useful. Apparently from time to time I take things for granted too quickly, and at the time the question seemed rather easy (I thought the implied meaning was, that when taking the limit for a massless object going to the speed of light, it's rest energy becomes infinite, referred to as becoming "infinitely heavy", etc. -- which by the way seems to me typically like a homework question, another reason for my hasty assumption). Reading Bernhard's last post, I realize the question is not nearly as trivial as I at first thought, and I probably misunderstood it the first time as well.

One of the dangers of communicating through written text... you never know who's on the other side of the line, do you. In this case, I think I'd better shut up because if you're right Pete, there's probably more I can learn from Bernhard than the other way around
Bernhard, I hope I haven't offended you (and if by accident I did, that you will accept my apologies).

11. Nov 17, 2007

### bernhard.rothenstein

kinetic energy in special relativity