# Why can light travel at the speed c?

1. Jul 2, 2011

### mibaokula

hey guys, i'm new to this forum so i dunno if this has been already asked before

i saw this equation on special relativity explaining why objects with mass cannot travel at the speed of light

E = mc^2/(1-(v^2/c^2))^0.5
( hard to write on a computer. view the attachment for the equation or link here: http://upload.wikimedia.org/math/2/4/0/24090a520815f2d76b2be996acc6c9e2.png)

this makes sense right; if an object travels at the speed of light, the denominator becomes 0 and if the object has mass (say for example one arbitrary unit) you get a large number divided by zero - in other words an undefined energy to make this happen - impossible
but what confuses me it that if you take a photon which can travel at c, you still get an undefined energy. my question is: how are photons able to travel at c in a vacuum? do they carry undefined amounts of energy? or does the equation not work for light?
thanks

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2. Jul 2, 2011

### Staff: Mentor

That's it. You need to use this more general equation, which works for all particles, including massless photons:

$$E^2 = mc^4 + p^2c^2$$

(p is the momentum of the particle.)

3. Jul 2, 2011

### ghwellsjr

Actually, a large number divided by zero is not undefined, it's infinite.
In this case you are dividing zero by zero which is indeterminate but can be an exact number, you just have to solve the problem some other way.

4. Jul 2, 2011

### Drakkith

Staff Emeritus
I thought ALL numbers divided by zero are undefined.

5. Jul 2, 2011

### bcrowell

Staff Emeritus
If 0/0=x, then x should be a solution of 0=0x. All values of x are solutions, so any real number is a possible value of 0/0. That's why it's called an indeterminate form.

It's misleading to say that 1/0 and 0/0 are both undefined and just leave it at that, because 1/0 is infinite, but 0/0 is an indeterminate form.

6. Jul 2, 2011

### ghwellsjr

Consider A/B=C.

Then A=B*C, correct?

Now if A=0 and B=0, we have:

0=0*C, correct?

This equation is true for every value of C, correct?

Therefore, 0/0 can be any number you want.

Last edited: Jul 2, 2011
7. Jul 2, 2011

### HallsofIvy

Yes, that is why it is called "indeterminant". But "infinite" is included in "undefined".

8. Jul 2, 2011

### bcrowell

Staff Emeritus
I'm not saying it's wrong to call it undefined, just that it's misleading. It hides an important distinction.

9. Jul 2, 2011

### Pengwuino

The past few posts have given me a little confusion because when you're talking about a photon, you're separately taking the limit of $m \to 0$ and $v \to c$ but one is not dependent on the other. If I could decide I simply want to take m to 0, you should find the first term drops out and end of story, it's 0, not infinite.

10. Jul 2, 2011

### atyy

Can you explain more? I have never understood this point. It was always explained to me that the relativistic mass is a good heuristic for showing that only massless particles can travel at the speed of light. I have never found the explanation that the true equation is E=rest mass+momentum2 to be any more rigourous, since momentum is not defined in the same way as for massive particles anyway. The only way I know to make it "correct" is to invoke quantum theory and say that mass is ultimately a dispersion relation of a wave. It'd be cool if a beloved 0/0 heuristic has more to it than I learnt.

11. Jul 2, 2011

### PAllen

I'll reiterate my favorite solution, which I haven't seen any book. I makes no use of division by zero, no funny definition of momentum.

For any particle with mass, it is straightforward to derive that:

sqrt(1 - v^2/c^2) = 1 - KE / (KE + E0)

If you want to have a particle with E0=0 and KE > 0, and have it follow as many of the same rules as normal particles as possible, then v=c is immediately implied. No limit, no zerodivide, no alternate momentum definition. THEN, having established this, you use :

E2=mc^4+p^2c^2

to arrive at p=E/c, where we know that for massless particles all E is kinetic, and m=0.

12. Jul 2, 2011

### Passionflower

If 1/0 = infinite then 1/infinite - 1/0 = 0 but 1/infinite is undefined so it could not be true.

13. Jul 2, 2011

### Matterwave

The limit as x approaches 0 for 1/x doesn't even exist since the limit approaches plus and minus infinity depending on which way you go, so how can you define 1/0 as infinity? Defining it as negative infinity would also be equally valid.

Defining 1/0 to be infinity causes all sorts of problems. It would define, then, infinity*0=1. But infinity*0 is an indeterminate form.

Also, defining 1/0 to be infinity would first require the definition of infinity to be one of the integers. This definition would ruin the fundamental theorem of arithmetic since infinity could not be uniquely factored into primes.

14. Jul 3, 2011

### ghwellsjr

The OP was asking about an equation for the energy of a moving particle. He stated that a massive particle has an undefined energy when moving at the speed of light and said that was impossible. I've never heard of anyone stating that the energy of a massive particle becomes undefined when traveling at the speed of light. I've always heard that it takes an infinite amount of energy to accelerate a massive particle to the speed of light.

Then he stated that a massless particle (a photon) also has an undefined energy and he wondered if photons therefore have undefined amounts of energy or if the equation doesn't work for light. But the equation doesn't yield an undefined energy for light in the same way that it does for a massive particle, rather 0/0 is indeterminate, it can have any value, you just have to determine that value some other way. It's not that the equation doesn't work for light in the sense that it is no longer true, but rather it just doesn't tell you which of the many possible answers is the correct one.

15. Jul 3, 2011

### ghwellsjr

I'm not following this at all. If 1/0 is infinite then 1/infinite = 0. Where did you get - 1/0? And why do you say that 1/infinite is undefined?

16. Jul 3, 2011

### mibaokula

how would this equation work in terms of particles like the tachyon which moves at superluminal speeds (assuming it exists - which it probably doesn't)

17. Jul 3, 2011

### Staff: Mentor

It still works for tachyons, but E, p, and m are all imaginary numbers.

18. Jul 3, 2011

### mibaokula

awesome, thanks again for the quick reply. i am somewhat enlightened.
although i'm sure that energy cannot be imaginary.

19. Jul 3, 2011

### Staff: Mentor

But you're OK with imaginary momentum and mass?

20. Jul 3, 2011

### mibaokula

i guess you're right. its all crazy!!
i suppose negative or imaginary mass would sum up to gravity having a repulsive effect or something. i guess that's feasible. if mass is imaginary then using any momentum equation:
p=mv or planck's constant/lambda; i suppose the momentum also is imaginary.
but the energy works out to be imaginary too (according to your equation) and that surely cant be right?
if you slot the numbers into the original equation i had before, you get a complex number divided by a complex number - a real number. how does that work??
surely tachyons cause the equations to lose their meaning - my main reason for thinking they don't exist

21. Jul 3, 2011

### mibaokula

try typing 1/0 into a calculator - its undefined. if you think about it this way, even if you had an infinite number of "zeros" you cannot get 1 - hence its not infinity but undefined.

likewise with 0/0, you might think that the answer is infinity as the other guys said there are many answers - infinite answers not just infinity. 1 x 0 is 0. 2 x 0 is 0. 3 x 0 is 0 and so on using this same logic.

so back to my original question, although its been answered, using that particular equation, photons must have an undefined amount of energy

22. Jul 3, 2011

### mibaokula

i think (if einstein's special relativity is right), they are dependent on each other. for any object to travel at c m/s, it must have no mass. so if you plug the numbers in the equation, you get 0/0 - which sure is confusing!

23. Jul 3, 2011

### atyy

In my understanding, the idea that massless particles can travel at the speed of light isn't a purely classical idea. In classical physics, light is a wave, not a particle. So the 0/0 thing is a "heuristic".

With quantum physics, everything becomes a wave, and the "mass" of a particle is seen to be a property of a wave. In this framework, since some waves have "mass", then light can be massless. However, there is no 0/0 - what this means is that all frequencies of light travel at the same speed. A massive particle is a wave in which different frequencies (energies) travel at different speeds.

24. Jul 3, 2011

### mibaokula

dunno about that but in the standard model, the "photon" has a mass of zero. so i kinda put that in the equation. i do not believe any electromagnetic wave has a mass although they can have momentum depending on their wavelength

25. Jul 3, 2011

### atyy

A massless photon simply means that all frequencies of light travel at the same speed. The equation is not strictly valid for massless particles, although it is suggestive that such particles can exist.

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