0
$\begingroup$

I have had this question for quite some time: As I understand it, in string theory there is tension along the strings such that this tension is between different elements of the string in space. Thus, a string will move in spacetime as if different elements of it were connected in the target space. However, this does not by itself mean that such elements will be connected through time (according to the target space's local definition of time). Now, if we take special relativity (SR) somewhat literally, then time ($ct$) and space (say, $r$) should be treated equally and we may argue that time, in the form of $ct$, should be able to be thought of as a length. Note that I am not referencing the use of SR transformations but the idea that $ct$ can be thought of as a length which seems to be standard in many treatments of relativity.

So my question is: could there then exist strings that have tension between different elements in the time defined locally in the target space? If this were the case, then, using the local coordinates of the target space as $(r,t)$, there might be tension between one particle appearing at $(r,t_1)$ and another particle appearing at $(r,t_2)$ if $t_2-t_1$ were sufficiently small and both particles were a part of the same string. (I realize that this suggests some unusual phenomenon, but if we consider the expected small scale of strings I'm not sure that these phenomena can be ruled out conceptually.) If not, then why not?

I will attempt to clarify my meaning as requested. (I'm not sure how to use the MathJax equations on here just yet, so I have tried to use as little math as I could.)

Improve this question
$\endgroup$
4
  • $\begingroup$ The whole modern physics takes STR literally and STR does not say, that time and space should be treated equally. They are not and cannot be treated equally at all. $\endgroup$
    – Umaxo
    Nov 30, 2020 at 10:02
  • $\begingroup$ SR leads to Minkowski space and Minkowski space is ingrained with the idea that space and time are treated on an equal basis - they combine to form space-time which is a 4-dimensional manifold. I believe you are referring to space-like versus time-like vectors which are indeed different. I am not referring to space-like and time-like, but to space (r) and time (ct). $\endgroup$
    – PrawwarP
    Nov 30, 2020 at 10:09
  • 4
    $\begingroup$ You probably know this but, just incase you're not familiar, strings are paramaterized by two numbers, usually $\sigma$ and $\tau$ which are spacelike and timelike, with $\sigma \in [0, \pi]$. These are (local) coordinate in the worldsheet. The string, in $D$ dimensional Minkowski space, is then written $X^{\mu} (\sigma, \tau)$, where $\mu$ is the usual spacetime index. The $\tau$ paramaterizes the trajectory (so they move in a timelike direction) along the worldsheet. Maybe you could clarify the question? $\endgroup$
    – Eletie
    Nov 30, 2020 at 10:13
  • $\begingroup$ @Umaxo Quantum field theory (QFT) is premised upon the idea that SR says that space and time should be treated on an equal basis. Note that this does not say that the nontrivial signature of the Minkowski metric is insignificant, but that the actual coordinates must both be parameters. Additionally, looking at "Relativity: Special, General, and Cosmological" reveals that, when used in SR, Minkowski space most definitely involves space and time definitions. $\endgroup$
    – PrawwarP
    Nov 30, 2020 at 10:21

1 Answer 1

Reset to default
2
$\begingroup$

This is a long comment.

This is from the wiki article on strings:

In physics, string theory is a theoretical framework in which the point-like particles of particle physics are replaced by one-dimensional objects called strings. String theory describes how these strings propagate through space and interact with each other. On distance scales larger than the string scale, a string looks just like an ordinary particle, with its mass, charge, and other properties determined by the vibrational state of the string

So this :" and both particles were a part of the same string" is a fundamental misunderstanding of string theory. Each elementary particle of the standard model is a string .

Different particles are different strings on the appropriate vibration level.

You are confusing classical strings with the quantum mechanical strings of "string theory". Tension is an attribute of classical strings, which even within a special relativity frame are not the strings of string theory.

Improve this answer
$\endgroup$
8
  • $\begingroup$ So the string tension in the Polyakov action is not a tension in the sense of two events being connected through space? Also, how do the strings then represent interconnected events (or is this also a misunderstanding on my part)? $\endgroup$
    – PrawwarP
    Dec 1, 2020 at 1:29
  • $\begingroup$ At the quantum level there are no forces,, of the type F=ma, and tension is a classical force. People use the term Force, because one needs a vocabulary to discuss models for the dp/dt exchanged in elementary particle interactions , the momentum transfer at Feynman vertices ( shorthand calculation help) . String theory model developed to go to smaller distances and higher energies , and solve problems with the QFT of the standard model. $\endgroup$
    – anna v
    Dec 1, 2020 at 5:57
  • 2
    $\begingroup$ Strings do not represent interconnected events, in the same way that particles do not represent interconnected events. The interconnection is seen when Feynman diagrams are drawn and many particles are involved.There are the corresponding Feynman type diagrams for strings calculations, and each particle represented now by a string. $\endgroup$
    – anna v
    Dec 1, 2020 at 5:57
  • $\begingroup$ So the "tension" you quote of a string is within the representation of a single particle, $\endgroup$
    – anna v
    Dec 1, 2020 at 6:03
  • 1
    $\begingroup$ I think that the treatment of space and time you envisage depends on Lorentz invariance and the algebra of the four vectors. "Moves" as a concept needs a special position for time. so maybe you mean makes a corresponding pattern. If the algebra is corretctly done I do not see why not, but it could not ,but there would be no way of seeing it experimentally, in my opinion.( I am an experimental physicist and my knowledge of the details of mathematics of string theory is) minimal. $\endgroup$
    – anna v
    Dec 1, 2020 at 10:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.