Is it right to assume that in 4-dimensional spacetime the speed of any object is constant? I mean a stationary object travels straight to the future but any spatial movement simply means that the direction of the speed vector doesn't point straight to the future. Because that's how I roughly understand what happens with the time flow when the object travels at relativistic speeds. But this would mean that for an object traveling spatialy at light speed the speed vector would be perpendicular to the time direction, so the speed in 4th dimension (time) would be 0. I'm sorry if it's stupid. I'm not a physicist but I think about these things a lot and I'd appreciate correcting my way of thinking. I do understand that motion is relative. Yet it bugs me that if I was right, then photons travelling at light speed would effectively be frozen in time.
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1$\begingroup$ then photons travelling at light speed would effectively be frozen in time. Light/photons do not experience time. They follow what’s called “light-like” or “null” geodesics. That is, the distance it travels from one point in spacetime to another is “zero” and it has no proper time (and this is true for all massless objects). The question about light and time has been asked here many times, so you may want to try the site search tool above for detailed information. $\endgroup$– joseph hCommented Feb 17 at 23:33
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$\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$– Community BotCommented Feb 18 at 0:01
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$\begingroup$ @josephh that's more or less my point. The photons don't experience time. They can't because they always travel perpendicular to time. $\endgroup$– ElmoVanKielmoCommented Feb 18 at 0:05
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$\begingroup$ @Community I want to understand if movement through time and space can be represented uniformely using a vector with constant value but varying direction. $\endgroup$– ElmoVanKielmoCommented Feb 18 at 0:08
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$\begingroup$ "The photons don't experience time. They can't because they always travel perpendicular to time." The path of a photon is perpendicular only to itself. $\endgroup$– WillOCommented Feb 18 at 18:56
2 Answers
In Minkowski space, there are three kinds of intervals, or separations between points, e.g. they are "time-like", "space-like" and "null". Light travels along null lines, i.e. lines which have 0 distance between any two points on the line. Such null paths are not perpendicular to the time-like paths. For a stationary object, (purely temporal direction) its path or "world line" is always at a 45 degree angle to null paths.
There are no normal physical objects which posses a purely space-like path, paths which are perpendicular to purely temporal paths, because such paths require infinite speed. However, not all space-like paths require infinite speed. All such paths do, however, require speeds faster than light, and are thus not possible for normal physical masses (hypothetical particles called tachyons reside in this domain).
Normal physical masses must stay in the realm of stationary or less-than-light speeds, a hyper-dimensional set of space-time points known as the "light-cone". All paths in this region are time-like. Navigation through space time is thus along time-like paths which have velocity vector that may have varying magnitude (from 0 to approaching c) and direction (purely temporal to approaching 45 degrees from temporal).
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$\begingroup$ It is worth noting that in 4 dimensional space time, the magnitude and direction of a velocity vector are not completely independent, e.g. the angle from the purely temporal depends on the magnitude itself such that the angle is zero when the magnitude is 0 and approaches 45 degrees as the magnitude approaches 0. One cannot, e.g. have a mass that is traveling with non-zero magnitude in the purely temporal direction, such paths are time-like, but not purely time-like. $\endgroup$ Commented Feb 18 at 13:53
But this would mean that for an object traveling spatially at light speed the speed vector would be perpendicular to the time direction, so the speed in 4th dimension (time) would be 0.
You need to be aware that in relativity there are two kinds of time, coordinate time and proper time. In a normal Minkowski spacetime diagram we use coordinate time for the vertical axis and in these diagrams, light always travels at 45 degrees to the vertical axis or at a rate of one unit of distance per unit of coordinate time.
Coordinate time is measured by a set of clocks that are stationary in a given reference frame and the time interval between spatially separated events uses two separate clocks to calculate the time interval.
Proper time can be thought of as the time that is measured by a moving clock that is transported from one event to the other. Proper time for a photon is slightly confusing, because no physical clock could be transported at the speed of light so it is hypothetical. This is the type of time I think you are thinking of and in a graph of distance versus proper time on the vertical axis, light would indeed be travelling perpendicular to the proper time axis.
Proper time can be thought of as the rate of ageing of a particle. The fact that photons do not experience proper time means they do not age. This is supported by the fact we still see photons in the Cosmic Microwave Background that were emitted in the very earliest stages of the universe. Photons live for ever or at least until they interact with something.
@Community I want to understand if movement through time and space can be represented uniformly using a vector with constant value but varying direction.
On a chart with distance versus proper time, the velocity in terms of proper time is called the celerity and for a photon the celerity is infinite. On such a chart, there is a sense that you could say that a vector representing the distance travelled per unit coordinate time has a constant magnitude and a varying angle.
Yet it bugs me that if I was right, then photons travelling at light speed would effectively be frozen in time
Photons are not frozen in time in the normal sense as they move relative to normal coordinate time, but you are right that they do not experience proper time and do not age.
Particles without mass cannot remain stationary with respect to the spatial axes and particles with mass cannot remain stationary with respect to the proper time axis.
