Tag Info

New answers tagged

0

Galilean relativity does not automatically hold in Newtonian mechanics. You are correct that Newton's first two laws appear at first sight to be invariant under galilean transformations, but a galilean transformation only transforms spatial coordinates,they don't actually tell you how forces transform. If your force laws just has forces that depend only on ...


2

Let's consider a hover craft or a rocket that will accelerate up to a certain elevation and keep hovering at it, as a better vehicle for our purposes than a plane. Let's forget about air momentarily. What will happen is that the moment our air craft leaves contact with the ground, its initial tangential velocity will be the same tangential velocity of ...


1

Where is the flaw in my thinking? In your concept of time. It's little more than a cumulative measure of local motion, see A World without Time: The Forgotten Legacy of Godel and Einstein]. Your macroscopic motion relative to some other guy results in you measuring his local motion to be slow, whilst he measures your local motion to be slow. This sounds ...


4

To measure time, a duration, you need two moments – when you press "start" and "stop" button on the stopwatch, respectively. But because the two objects are moving relatively to each other, it isn't possible for them to "meet" at both moments. If their locations coincide at the "start" moment, for example, so that their clocks may be compared at this "start" ...


0

To answer my question #1: As Demosthene said, $Λ^μ_ν$ depends only on the relative velocity between frames $S$ and $S'$, but not the relative velocity between particles. This allows me to make the assumption, which I numbered 6 in my initial question, thus solving my problem.


0

EM waves are mass-less. The time dilation effect you mention only applies when a travelling body has mass. Note that for a body that has mass it would take an infinite amount of energy to travel at c.


0

you are correct that the EM field propagates at c, but as the comment from Acuriousmind mentioned, there is no frame of reference traveling at that speed. What that means in simpler words, is that the equations break or diverge for anything traveling at the speed of light. Light is not bound to time dilation or length contraction for that same reason. Only ...


0

Electromagnetic waves rely upon oscillations to propagate Electromagnetic waves are propagating disturbances (or oscillations) in the electromagnetic field and the electromagnetic field exists everywhere and everywhen. But the electromagnetic field itself does not 'travel' at $c$ so your reasoning isn't clear to me.


-1

I am sorry, but I have no time to find your mistake for you. However, I can give you a tip that I am 100% sure would work. Your mistake can be very easily found by dimensional analysis. Start from the beginning of your derivation and check all formulas. The first one which makes no sense due to dimensions is wrong. You can just say that because $c$ is one ...


1

You work too hard and the idea of setting $c=1$ may make problems. I copy here the equations that you obtained and that I found correct. So, 1. The speed of light is the same in each frame implies $$Act + Bt = c(Dct + Et), \overset{(x=t)}{\implies} Ac + B = Dc^2 + Ec. \tag{i}$$ Reversing the frames gives indeed $$B = -Ev. \tag{ii}$$ Also in the frame ...


3

Both the the Earth and the Sun orbit around the solar system barycentre. This is defined as the centre of mass of all the bodies in the solar system. Because the Sun contains the vast majority of the mass of the solar system then the barycentre is very close to the Sun. The picture below, from the wikipedia entry on the solar system barycentre, has the ...


1

There are at least 8 more planets in the solar system, besides the Earth (and some more were discovered). When Copernicus decided to place the Sun in the center of the solar system, instead of the Earth, that was mainly because this arrangement simplified drastically the form of the orbits of the other planets. With the Geocentric model of the solar system ...


3

In binary systems, each object is so affected by the others gravity that they have significant orbit. The sun has so much inertia that the earth's pull barely moves it, but the earth certainly revolves around the sun. In the reference frame of the Earth however, the Sun does revolve around the Earth.


0

There is no "correct" speed. All speeds are relative to the observer, but this does not mean you can watch something moving faster than light even if you are moving one way at 0.6c (60% the speed of light) and someone else is moving the other way at 0.6c. Even though typical logic would dictate that you see each other move at 1.2c (20% faster than the speed ...


2

Short answer : the same light goes at the same speed (c) relative to any observer. There is no grid. This is counter-intuitive : if you're standing in a bus traveling at 30 mph and you walk at 3 mph towards the driver, you walk at 30 + 3 = 33 mph relative to the road. But that doesn't work for light : if the bus travels at c/2 and you shine a flashlight ...


3

Suppose we play a racing game. I scatter a little bit of dust around space, then you come by me in your spaceship at some speed $v$. Let's start with $v = c/2$, just so we're not contentious. Right as you pass, I fire a really bright laser pulse in the direction you're going. You're racing the laser light. The dust means that you see reflections of it, so ...


2

It's relative to all inertial reference frames--in special relativity the coordinates of one inertial frame are related to the coordinates of another by the Lorentz transformation, and this transformation has the property that anything with a coordinate speed (change in coordinate position divided by change in coordinate time) of c in one inertial frame will ...



Top 50 recent answers are included