1. Mar 10, 2013

### Dan248

Hi.
I have some questions that I'm not sure about to do with this equation. I may just be missing something obvious but would be nice to get an answer nonetheless..

First is to do with the LHC. My understanding is that they are colliding particles at such force that they break down into particles that can't be observed as they are without mass. But the energy needed to make this happen is immense, yet the result has no mass. Doesn't this defy the fact that energy=mass?

The other question is... There is no state of rest, all is relative. So when we measure the speed of light in a vacuum, it's still relative to the speed at which we move... The closer you are to the speed of light, the slower you'd see light as travelling, so if you were to use e=mc2 on your own mass, then the speed of light will have reduced relative to your increase in energy, so the result would come out exactly the same? Or am I missing something really obvious here....

2. Mar 10, 2013

### Von Neumann

Hi,

I'm not sure what you're asking in the second part, but the speed of light is a universal constant. If you are moving, say "with" a beam of light (side-by-side), it will appear to be moving at a speed of 3.0x10^8 m/s regardless of your relative speed. The constancy of the speed of light is a foundation in special and general relativity, and has profound implications.

Hope that helps you out!

3. Mar 10, 2013

### phinds

No, the speed of light is the same regardless of your motion and/or the motion of whatever emits the light. In the sense in which you are talking, you will always measure the speed of light as c.

4. Mar 10, 2013

### Staff: Mentor

This one belongs in the relativity forum, I think.... Likely enough a moderator will move it.

The speed of light in a vacuum is the same for all observers. Say I'm at rest, and at time zero I send a flash of light off in some direction and at the same time you take off on a spaceship traveling at .25 the speed of light in the same direction.

Relative to me, you are moving at .25c, the light is moving at c, and the distance between you and the light is increasing at .75c. Relative to you, the light is moving away from you at c, I'm moving on the opposite direction at .25c, and the distance between me and the light is increasing at 1.25c.

5. Mar 10, 2013

### Staff: Mentor

Energy always has mass given by Einstein's equation. When someone speaks of a particle "without mass", they mean "without rest mass", meaning that all the energy and hence mass comes from the kinetic energy of the particle's motion. (BTW, these particles are the only ones that can move at the speed of light, and that is the only speed at which they can move).

There's no problem observing massless particles... Your eyes do it by the trillions every second when they detect light.

Last edited: Mar 10, 2013
6. Mar 10, 2013

### DennisN

Hi Dan, and welcome to PF!
No, this is not correct (except for annihilation). There are many particles produced in colliders which have mass.
No. As pointed out above, the speed of light is constant according to Special Relativity (see point #2).

Last edited: Mar 10, 2013
7. Mar 10, 2013

### Dan248

Apologies if these are really basic questions, but it's hard to find answers and information on this sort of thing, so I appreciate the answers...

In this situation, if you're at rest, how do you observe the light moving away from you at 1.25c? Isn't it then moving away from you faster than the speed of light?

8. Mar 10, 2013

### Staff: Mentor

I don't observe it moving away from me at 1.25c, I observe it moving away from me at c while you are moving in the same direction at .25c.

You, however, see me moving to the left at .25c and the light moving to the right at c.

No one sees anything moving at a speed greater than c and everyone sees the light moving at c.

9. Mar 11, 2013

### Dan248

So If I am moving at 80% the speed of light, observing a light beam that has originated from the same location as me, do I see the speed that it is moving away from me as C or do i see it as 20% of C???

10. Mar 11, 2013

### Staff: Mentor

You will measure the speed of light as c. Everyone does!

11. Mar 11, 2013

### Dan248

So isn't Light's speed effectively infinite in that case?

12. Mar 11, 2013

### Staff: Mentor

Well, no. Light's speed is effectively c. The tricky part is that it's c with respect to any observer, regardless of that observer's motion.

Why would you say it was infinite?

13. Mar 11, 2013

### ebodet18

Okay I have a question, E=mc^2 only works with 100% efficiency with anti matter as it's energy source. How do we get anti matter? can we?

14. Mar 11, 2013

### phinds

Anti matter is created routinely in particle accelerators, but only in tiny amounts.

15. Mar 11, 2013

### phinds

Which part of post #3 did you not understand?

16. Mar 11, 2013

### Pyzyq

dude thats not how light works. relativity states that light is special unlike other things so when you move at 80% off the speed of light, you see the light pass you at the full speed of light, about three-hundred million meters per second. unlike when u r in a car and you are going 60 and the guy going 70 passes you at 10 miles per hour. light. is. special.

17. Mar 11, 2013

### Pyzyq

yes you are missing alot of things first of all particles are made out of the most fundemental constituants of matter called quarks (forgive my horrible spelling), the quarks cant be seperated so they cant be studied on their own. Hadrons: protons neutrons, and electrons have three quarks. they crash various hadron particles in various different types of experiments at .9c or almost the speed of light. the result is a bunch of other particles from other fields get nocked out of place and into existance ( not necisarilly existance) but into our plane of reference the faster we collide these particles, the more particles we get in result, we collided particles so fast we nocked loose a higgs boson which was really hard because it was so heavy and deeply embeded in the higgs field. you are confussed by the field of quantum electrodynamics, as you know, light is a massless particle called a photon when photons bounce off of the particles from the collision at the lhc they move crazy

18. Mar 11, 2013

### e.chaniotakis

Hey Dan!
I suggest that you chekcked some introductory relativity textbook (Check the general physics of Serway, Young etc). The law of velocity addition changes when u->c.
This way, if you approach an object that moves with c, and your velocity is .25c then you will see it moving with c again.
Same goes if you move away from it.
All this is a conclusion of the fact that all laws of physics are invariant for observers moving with constant velocity. A.k.a an observer moving with u and another moving with v= u+u' measure the same physical laws.
One set of these laws are the Maxwell equations. If you go from one reference frame to another moving with different velocity, the electromagnetic wave ecquation will be the same -- (d^2/dx^2)E =(1/c)^2*(d^2/dt^2)E -- the 1/c term remains invariant and this is the 1/speed of light.
Thus speed of light must be invariant among 2 observers with constant relative velocity and thus they will both measure c no matter what.

19. Mar 12, 2013

### DennisN

No, this is not correct (you might know this, but wrote to quickly ). Electrons are not hadrons and they are not made of quarks, electrons are elementary particles.

For Dan and others:

Elementary particles are particles with no known substructure, that is, the quarks, leptons (e.g. electrons) and the fundamental bosons (including all of their corresponding elementary antiparticles). Hadrons (baryons, mesons) are particles made up of quarks. Baryons (e.g. protons, neutrons) are made up of 3 quarks. Mesons (e.g. pions) are made up of 2 quarks. Welcome to the particle zoo . For an overview of the zoo, see this link.

20. Mar 12, 2013

### uperkurk

It's an extremely difficult concept to grasp and many people struggle with it including myself. If you and a particle of light were to have a race on a race track. You're running along side the particle of light at 99% the speed of light, that particle will appear to be pulling away from you and gaining distance on you at C.

The people in the crowd watching see you and the light particle almost tied for the lead.

If you're sitting in a plane flying at 99% the speed of light and you lean out the window and fire a particle of light, that light will gain distance on you @ about 186,000mph. While the people on the ground watching will see the light barely making any distance on you.

21. Mar 12, 2013

### HallsofIvy

In Gallilean physics if you are moving, relative to the ground, at speed u and you fire or throw an object ahead of you, relative to you, at speed v then its speed, relative to the ground, is u+ v.

In relativity, that is only approximately correct (the lower the speeds the better the approximation). The correct formula is $(u+ v)/(1+ uv/c^2)$. In particular, if other people see you moving with speed v and you shine a beam of light ahead of you with speed, relative to you, c, the observers will see it moving with speed, relative to them
$$\frac{v+ c}{1+ \frac{vc}{c^2}}= \frac{v+ c}{1+ \frac{v}{c}}= \frac{(v+c)(c)}{(1+\frac{v}{c})c}= \frac{(v+ c)c}{v+ c}= c$$.

In other words, all observers, whatever their speeds relative to you, will see light moving with speed c.

22. Mar 12, 2013

### ghwellsjr

You have to be careful with the terminology of "seeing light traveling". We can't see light traveling like we can see material objects traveling. We use light to see with, don't we? We shine a light on an object and then we can watch it move but how can we do the same thing with light? We can't, can we? So what do we do instead? We put a material object out some measured distance from us and we send a flash of light to it and we "see" the reflection when it gets back to us. We then measure how long it took from when we sent the flash until we detect the reflection and we divide that in half and use that to calculate the round-trip speed of light by dividing that into the measured distance away the object is from us. Whenever any inertial observer makes this measurement, no matter what their state of motion, as long as the object is also inertial a fixed distance away, they will get the same answer for the speed of light, the universal constant, c.

But how can we tell if it took the same amount of time for the light to get from us to the object as it did for the reflection to get from the object back to us? We can't. It's impossible to know if it takes the same amount of time. So what Einstein did in his second postulate is define those two times to be equal.

Once we do that, we can always say that in any Inertial Reference Frame (IRF), light travels at c in all directions. It's not because we measure it to be so, it's because we defined it to be so.

Let me show you how this works in an IRF in which the observer is stationary. In this diagram, he is shown as the blue line with dots every nanosecond and the reflecting object is shown as a red line 6 feet away. (I'm using the speed of light to be 1 foot per nanosecond.) At time 4 nanoseconds, the observer sends a pulse of green light to the object which travels at c (along a 45-degree angle) which takes 6 nanoseconds and then the reflection takes another 6 nanoseconds and so he detects the reflection at his time of 16 nanoseconds according to his clock. He then subtracts the emitted time of 4 from the reflected time of 16 to get 12 and divides that by 2 to get 6 nanoseconds and then divides that time into the distance of 6 feet to get 1 foot per nanosecond as the speed of light:

Now let's see what happens if he does the same thing in an IRF moving at -0.6c with respect to the original IRF:

Now you will note that as far as the observer is concerned, his measurement of the round trip speed of light comes out exactly the same even though the light clearly takes a lot longer to get to the object than it does to get back to him. But he can't know that this is what is happening. Why is this? It's because in this frame his clock takes longer to tick out each nanosecond (Time Dilation) and because his ruler is Length Contracted from 6 feet to 4.8 feet. (Gamma at 0.6c is 1.25.)

So please understand that whenever anyone says that they measure the speed of traveling light to be c or whenever they say that they see light traveling at c, they aren't talking about an actual measurement or an actual observation, they are talking about the definition of traveling light according to Einstein's definition of Special Relativity.

Hope this helps clarify any confusion.

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23. Jan 18, 2016

### David Lewis

Pyzyq wrote: " ...relativity states that light is special unlike other things so when you move at 80% off the speed of light, you see the light pass you at the full speed of light, about three-hundred million meters per second. unlike when u r in a car and you are going 60 and the guy going 70 passes you at 10 miles per hour. light. is. special."

David Lewis wrote: I think the velocity addition formula is the same for all objects at all speeds. At speeds much less than c, however, simple arithmetic is a close approximation.

ghwellsjr wrote: "...whenever they say that they see light traveling at c, they aren't talking about an actual measurement or an actual observation, they are talking about the definition of traveling light according to Einstein's definition of Special Relativity."

David Lewis wrote: In the 2nd diagram, the green light travels 12 feet in 12 ns to get to the reflector, and then takes 4 ns to travel 4 feet on the return trip.

Last edited: Jan 18, 2016
24. Jan 18, 2016

### Staff: Mentor

With the proviso that "all speeds" means "all speeds less than or equal to $c$", yes, this is correct.

(Btw, you can use the quote feature to quote portions of posts that you want to respond to. If you highlight what you want to quote, two buttons should pop up, "Quote" and "Reply". I usually use the latter, it just pastes the quote into the edit window you use to compose your post.)

25. Jan 19, 2016

### Battlemage!

Wouldn't having two different speeds for the one way speed of light and reflected speed of light violate the conservation of momentum? For example, if you had an infinite one way speed and a .5c reflected speed, the total trip would have an average speed of c. p = E/c, right? So on the way there you would have a momentum of zero, but on the way back you would have a momentum of 2E/c, which means momentum is not conserved. That seems to break physics, doesn't it?