What is the 4-velocity of an object?

[epr2008]

My biggest qualm with the special theory of relativity and, for that matter, the general theory of relativity is the fact that both assume a little bit too much. My first area of skepticism is that so many cling to these as much as they do to Newton's laws, but there has been no tangible proof of either theory. While it seems to be very brilliant, it has also been proven logically and mathematically that beginning with a false statement can inevitably yield any result, and obviously even more so when that result is desired. This brings me to the foundation of the theories, the defined space-time interval given by

$$d{S^2} = d{l^2} - {(cdt)^2}$$.

My first question: Is there any proof that this is actually representative of "distance" in space-time? The theory clearly states that space and time are intimately connected, but this seems somewhat ill-posed as the definition of the interval obviously removes their dependence. I really do not know if this is a well-posed counterexample, but consider a stationary object in space with movement represented by a space-time coordinate system and let its position be the origin. Then after, say, one second for simplicity, the object has not moved and so

$$\begin{array}{l} 0 = {l^2} - {(ct)^2}\\ l = ct\\ l = 3.00 \times {10^8}{\rm{meters}} \end{array}$$

Which seems nonsensical since I stated that the object was stationary.

Next, I'd like to point out another assumption. Let $${l^2} = {x^2} + {y^2} + {z^2}$$ we have

$$d{S^2} = d{x^2} + d{y^2} + d{z^2} - {(cdt)^2}$$


Which brings me to my second question: Is there any proof that we occupy a 3 dimensional or even finite dimensional space, and is there even such a thing as "dimension"? For the former, it has become natural in mathematics that we represent spatial orientation by coordinates and coordinate axes, but this is merely a projection of actual space into an abstraction which we define. We can even project further into 2 dimensions and one dimension for special motions, and, because of transformational properties, orientation is preserved. But this still only says that these abstractions are projections and not actual representations. As for the latter, I do not think that there is a fathomable answer.

 

[Drakkith]

We have defined "Dimension" to be a specific thing, and according to what that definition represents, we live in 3 physical dimensions and possibly 1 of time as well. This is of course ignoring string theory and such.

Why does moving 3x10*8 meters in spacetime seem nonsensical to you? A "stationary" object is only stationary in the 3 physical dimensions, but not in time. Giving the distance in spacetime in Meters is merely convenience to a distance we already have. You could call it feet, inches, centimeters, or whatever you wanted to and it would all end up the same distance after conversions. Whether we KNOW we are traveling through spacetime or not, that's what we have defined it to be. If we say we travel X meters in spacetime per second while stationary, then we have simply called whatever distance it actually is, if any, X meters.

While starting with assumptions can lead to incorrect answers, remember that these theories have plenty of evidence to back them up. Thats the key here as it is everywhere in science.

Check here for some examples. http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/980327b.html

 

[Dale]

My first area of skepticism is that so many cling to these as much as they do to Newton's laws, but there has been no tangible proof of either theory.

This is complete BS. There has been more than 100 years worth of some of the most precise and painstaking tangible proof ever obtained in the history of science. I'm sorry, but this statement betrays complete ignorance on the subject. Here is a brief overview of the experimental results: http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

$$d{S^2} = d{l^2} - {(cdt)^2}$$.

My first question: Is there any proof that this is actually representative of "distance" in space-time?

Yes, see above

 

[Drakkith]

Interesting read Dalespam, thanks for the link. =)

 

[grav-universe]

I really do not know if this is a well-posed counterexample, but consider a stationary object in space with movement represented by a space-time coordinate system and let its position be the origin. Then after, say, one second for simplicity, the object has not moved and so

$$\begin{array}{l} 0 = {l^2} - {(ct)^2}\\ l = ct\\ l = 3.00 \times {10^8}{\rm{meters}} \end{array}$$

Which seems nonsensical since I stated that the object was stationary.

That describes a http://en.wikipedia.org/wiki/Spacetime between the events of two ticks of the stationary observer's clock. l is not 3 * 10^8 meters, it is 0 meters at the origin, and dS^2 is not zero, but -(3*10^8 m)^2. The space-time interval gives a relation between what observers in different frames measure between the same two events. For instance, the same space-time interval will also be measured between the same two events by a second observer moving with a relative speed to the first. The second observer measures the clock of the first observer to be moving through space, so adds a distance component to the space-time interval between the ticks, but also taking more time than one second according to the second observer's own clock, due to the time dilation observed of the first observer's clock, so dS works out the same as measured from any frame.

Next, I'd like to point out another assumption. Let $${l^2} = {x^2} + {y^2} + {z^2}$$ we have

$$d{S^2} = dx^2 + d{y^2} + d{z^2} - {(cdt)^2}$$


Which brings me to my second question: Is there any proof that we occupy a 3 dimensional or even finite dimensional space, and is there even such a thing as "dimension"?

This seems more a philosophical question, but in any case, they are called dimensions as we have defined that term. Each spatial dimension can be made at right angles to each of the others, and there are only three that can be measured. If others were somehow to be found, however, say other spatial dimensions a and b, then we would just add those to the space-time interval in the same way with

dS^2 = da^2 + db^2 + dx^2 + dy^2 + dz^2 - (c dt)^2

We can go ahead and add those now if we wish, but so far all objects' paths through the spatial dimensions a and b has been measured to be a=0 and b=0 as far as we can tell.

For the former, it has become natural in mathematics that we represent spatial orientation by coordinates and coordinate axes, but this is merely a projection of actual space into an abstraction which we define.

In a manner of speaking, this is correct. I personally would refer to it more as a "construction" rather than an "abstraction", however.

We can even project further into 2 dimensions and one dimension for special motions, and, because of transformational properties, orientation is preserved. But this still only says that these abstractions are projections and not actual representations.

That just means that the motions of objects along the other one or two spatial dimensions, for instance, is y=0 and/or z=0. The physical dimensions of y and z haven't gone anywhere, but we just don't need to consider them.

 

[PhilDSP]

Next, I'd like to point out another assumption. Let $${l^2} = {x^2} + {y^2} + {z^2}$$ we have

$$d{S^2} = d{x^2} + d{y^2} + d{z^2} - {(cdt)^2}$$


Which brings me to my second question: Is there any proof that we occupy a 3 dimensional or even finite dimensional space, and is there even such a thing as "dimension"?

If you are really interested in understanding these things you should probably start by studying the Maxwell Equations, especially how they were developed and later interpreted and restructured to the form we recognize today. The link between observational evidence of any sort and mathematical models has its basis there.

Mathematically, a "dimension" is a variable parameter that is not depend on any other variable parameter. The space dimensions are easy for us to visualize because we have early been taught geometry where distance is a common metric. Dimensions using a common metric must be defined to be orthogonal to each other or else the mathematics become invalid. Strictly speaking, the development of a common metric between time and the spatial dimensions is not an assumption but rather a matter of definition. That definition is implicitly built into the form of Maxwell equations which describe EM behavior that are used by both theorists and experimenters so naturally you see no deviations.

 

[yuiop]

This brings me to the foundation of the theories, the defined space-time interval given by

$$d{S^2} = d{l^2} - {(cdt)^2}$$.

My first question: Is there any proof that this is actually representative of "distance" in space-time? The theory clearly states that space and time are intimately connected, but this seems somewhat ill-posed as the definition of the interval obviously removes their dependence. I really do not know if this is a well-posed counterexample, but consider a stationary object in space with movement represented by a space-time coordinate system and let its position be the origin. Then after, say, one second for simplicity, the object has not moved and so

$$\begin{array}{l} 0 = {l^2} - {(ct)^2}\\ l = ct\\ l = 3.00 \times {10^8}{\rm{meters}} \end{array}$$

Which seems nonsensical since I stated that the object was stationary.

It is nonsense. For some reason you assumed dS=0 which is not correct. Since you stated the object is stationary then l=0 and the result for t=1 second is;

$$\begin{array}{l} dS^2 = {(ct)^2}-{l^2}\\ dS^2 = {(c)^2}-{0^2}\\ dS = 3.00 \times {10^8}{\rm{meters}} \end{array}$$

I have reversed the signs because it makes life easier, unless you are fond of imaginary numbers

Next, I'd like to point out another assumption. Let $${l^2} = {x^2} + {y^2} + {z^2}$$ we have

$$d{S^2} = d{x^2} + d{y^2} + d{z^2} - {(cdt)^2}$$

Here is another way of looking at it. The equation can written as:

$$c^2 d{\tau^2} = d(cT)^2 - d(x)^2 - d(y)^2 - d(z)^2$$

$$c^2 d{\tau^2} = d(x_1)^2 - d(x_2)^2 - d(x_3)^2 - d(x_4)^2$$

$$c^2 d{\tau^2} = d(x_n)^2$$

$$c = \frac{dx_n}{d\tau}$$

so the 4 velocity (displacement in all 4 coordinate dimensions per unit proper time) of any object according to any observer is always c.

Here the coordinate time multiplied by c and with a change of sign is treated just like any other 3 spatial coordinates. Some people say it is nonsense to talk of velocity in the time direction because velocity is a displacement divided by time and velocity in the time dimension becomes time/time. What they are forgetting is that we are talking about displacement in coordinate time, so that the velocity in the time dimension is (coordinate time displacement)/(proper time) and the velocity in the spatial dimensions is (coordinate spatial displacement)/(proper time). Velocity expressed this way is invariant when transforming to a different frame.