Physics Inquries

[Dreamerb90]

This thread is for any questions you may have about physics, not saying that ill answer them, I'm sure someone else will. Here are a few of mine. If you are sitting in a car, and another car passes you at 60 miles an hour, would it look slower than if you were going 60 miles an hour and the other car was passing at 60 miles an hour as well? According to the theory of relativity, nothing can go faster than the speed of light. However, if you throw a light bulb would it be the speed of light plus the speed you threw the light bulb? Post yours, I'm curious to see what goes on in your heads.

[Xei]

The speed of light is a fundamentally unchanging thing. It is the same for all observers, in all directions, no matter how fast the source is moving, or how fast the observer is moving. So no, if you throw a light bulb, from your perspective, light will still come out of it at the same speed. If you yourself were to run away from light beams coming from the sun, the light would still hit you at the same speed. If you ran towards the sun, the light would not hit you any faster. This single fact (it was first verified at the end of the 1800s) means we have to radically rethink space and time. To see one reason why, consider a train moving quickly along a track, and a person in the middle who sends a flash of light (from a laser pen for example) in both directions. Because light travels at the same speed for him, this is what he sees happen: The light hits both ends of the train at the same time. However consider somebody outside the train, watching it travel past them. This is what they see: Because light also travels at the same speed for THEM, the back of the train catches up with the light, and the light hits the back of the train before it hits the front. Neither of these facts are wrong; they are BOTH right. That is to say, for the man in the train, the light hits both ends simultaneously; for the man outside, the light does not. This means our idea of everybody having the same idea of what is simultaneous is total wrong. In 1905 Einstein gave us a new theory of space and time which explains things like this, namely the Theory of Special Relativity.

[Dreamerb90]

You've taken classes on this haven't you, your knowledge is impressive, all i know came from Carl Sagan's "Cosmos" Thank you for answering my question.

[Wayfaerer]

This is because time slows down and length contracts more for the light beam that passed you on the right relative to the light beam your currently passing and so partly traveling with, right?

[Puffin]

I can answer the cars question, albeit I'm not sure how well I'll be able to explain it. Here goes my ridiculously mathy explanation... Someone care to explain the actual physics of it? There's this quantity called a vector. Vectors have a magnitude (speed/size) and direction. Two or more vectors can be added up to create a single new vector. This is where directions come into play - for instance, if you have one car going at 60 mph towards an unmoving car, which is going at 0 mph, the perceived speed of the moving car (if you're in the unmoving one) will be 60 mph. Likewise, if you're moving at 60 mph and you're also driving towards a car moving at 60 mph, the two vectors add up to become 120 mph, the perceived speed of the car that's passing you. In other words, yes. If you're sitting still, the car will appear to go slower than if you were also traveling towards it.

[Wayfaerer]

The car situation can be described with vectors like Puffin said. Basically if your sitting still relative to the ground you only see the other cars velocity relative to the ground. If your moving relative to the ground too at a certain velocity, then you can think of yourself as stationary and the ground as moving passed you at that velocity. Add the velocity the other car is moving relative to the ground plus the velocity the ground is moving relative to you and you get the velocity you see the other car moving relative to you.

[Photolysis]

If you yourself were to run away from light beams coming from the sun, the light would still hit you at the same speed. If you ran towards the sun, the light would not hit you any faster. Though the light would be blue-shifted ever so slightly, of course. If you are sitting in a car, and another car passes you at 60 miles an hour, would it look slower than if you were going 60 miles an hour and the other car was passing at 60 miles an hour as well? Yes, the first would pass slower. It's trivial just to consider the velocities, which are a vector quantity. In the first case, when you're in the stationary car and another car passes at 60mph, the difference in velocities is obviously 60mph. In the second case, if you're going at 60mph in one direction, and are passed by a car going in the opposite direction at 60mph, the other car has a velocity of -60mph, which is a 120mph difference.

[DrunkenArse]

It's interesting that both questions are about "addition of velocities". In the classical regime, the formula is simple and you applied it to light and saw that it breaks down. The formula is simple. If an observer is moving with velocity v with respect to us and observes a particle moving with velocity u then we will see the particle moving with speed u + v. As you noted, this allows observers to measure light moving at any speed we desire. We know from electrodynamics that this is not possible. It is always measured moving at the same speed (in vacuum of course). This is the heart of the conflict between Newtonian mechanics and electrodynamics that Einstein resolved. The new formula for the addition of velocities becomes (simplified to the case where all motion is along the x axis) (u + v)(1 + uv/c2)-1 Note that this is just the old formula divided by a new term (1 + uv/c2)-1. Understanding how this term behaves will make everything clear. Note first that it's symmetric in u and v. This means that we can interchange the roles that they play without changing any numbers. Let's consdier the case when one of the velocities involved is 0. Say u. Then (1 + uv/c2)-1 = 1/(1 + 0v/c^2) = 1 and so we just recover the classical addition formula. We will measure the particle moving at velocity v + u = v + 0 = v. On the other hand, consider the case where u=c. Then (1 + uv/c2)-1 = (1 + cv/c2)-1 = (1 + v/c)-1. So our expression for the velocity that we will measure the particle moving at is (u + v)(1 + v/c)-1 = (c + v)(1 + v/c)-1 Now multiply top and bottom by c to get c(c + v)(c + v)-1 = c So we see that if a particle moves at the speed of light with respect to one reference frame, then it moves at the speed of light in all reference frames. Note also that when v and u are small then the term uv/c2 is very small and so the denominator is very close to 1 and the value will be very close to the classical form. Also, here's a cool video that touches on what Xei said but in video.

[Xei]

I came. The vid is hella cool, add it to sexy science.